Codeforces Round #310 (Div. 1) B. Case of Fugitive（set二分）

B. Case of Fugitive

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Andrewid the Android is a galaxy-famous detective. He is now chasing a criminal hiding on the planet Oxa-5, the planet almost fully covered with water.

The only dry land there is an archipelago of n narrow islands located in a row. For more comfort let's represent them as non-intersecting
segments on a straight line: island i has coordinates [li, ri],
besides, ri < li + 1 for 1 ≤ i ≤ n - 1.

To reach the goal, Andrewid needs to place a bridge between each pair of adjacent islands. A bridge of length a can
be placed between the i-th and the (i + 1)-th
islads, if there are such coordinates of x and y,
that li ≤ x ≤ ri, li + 1 ≤ y ≤ ri + 1 and y - x = a.

The detective was supplied with m bridges, each bridge can be used at most once. Help him determine whether the bridges he got are
enough to connect each pair of adjacent islands.

Input

The first line contains integers n (2 ≤ n ≤ 2·105)
and m (1 ≤ m ≤ 2·105)
— the number of islands and bridges.

Next n lines each contain two integers li and ri (1 ≤ li ≤ ri ≤ 1018)
— the coordinates of the island endpoints.

The last line contains m integer numbers a1, a2, ..., am (1 ≤ ai ≤ 1018)
— the lengths of the bridges that Andrewid got.

Output

If it is impossible to place a bridge between each pair of adjacent islands in the required manner, print on a single line "No" (without the quotes),
otherwise print in the first line "Yes" (without the quotes), and in the second line print n - 1 numbers b1, b2, ..., bn - 1,
which mean that between islands i and i + 1 there
must be used a bridge number bi.

If there are multiple correct answers, print any of them. Note that in this problem it is necessary to print "Yes" and "No" in
correct case.

Sample test(s)

input

output

Yes

2 3 1

input

output

No

input

2 1

1 1

1000000000000000000 1000000000000000000

999999999999999999

output

Yes

1

Note

In the first sample test you can, for example, place the second bridge between points 3 and 8, place the third bridge between points 7 and 10 and place the first bridge between points 10 and 14.

In the second sample test the first bridge is too short and the second bridge is too long, so the solution doesn't exist.

#include <bits/stdc++.h>

#define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it)

using namespace std;

typedef long long ll;

const int maxn = 2e5 + 10;

#define x first

#define y second

typedef pair<ll,ll> pll;

typedef pair<pll,ll> plll;

typedef plll Seg;

typedef pll Bridge;

bool cmpSeg(const Seg & a, const Seg & b)

{

	pll ta = a.x, tb = b.x;

	if(ta.y == tb.y) return ta.x < tb.x;

	return ta.y < tb.y;

}

Seg p[maxn];

Bridge a[maxn];

ll l[maxn],r[maxn],ans[maxn];

int main(int argc, char const *argv[])

{

	ios_base::sync_with_stdio(false);cin.tie(0);

	int n,m;

	while(cin>>n>>m) {

		for(int i = 1; i <= n; i++) {

			cin>>l[i]>>r[i];

		}

		for(int i = 1; i < n; i++) {

			p[i-1].x.x = l[i+1] - r[i];

			p[i-1].x.y = r[i+1] - l[i];

			p[i-1].y = i-1;

		}

		sort(p,p+n-1,cmpSeg);

		set<Bridge>Q;

		for(int i = 1; i <= m; i++) {

			ll a,id = i;cin>>a;

			Q.insert(make_pair(a,id));

		}

		set<Bridge>::iterator it;

		bool ok = (m + 1 >= n);

		for(int i = 0; i < n-1; i++) {

			if(!ok) break;

			it = Q.lower_bound(make_pair(p[i].x.x,0LL));

			if(it==Q.end()||it->x > p[i].x.y) {

				ok = false;

				break;

			}

			ans[p[i].y] = it->y;

			Q.erase(it);

		}

		if(ok) {

			cout<<"Yes\n";

			for(int i = 0; i < n-1; i++)cout<<ans[i]<<" ";

			cout<<"\n";

		}else cout<<"No\n";

	}

	return 0;

}

巴特西

Codeforces Round #310 (Div. 1) B. Case of Fugitive（set二分）

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